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2 years ago

What pressure will be exerted by 1.22 mol of CO2 ata temperature of 298 K and a volume of 500 mL?

Chemistry

1 answer:

kenny6666 [7]2 years ago

3 0

Explanation:

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Which is a diatomic molecule? A. He B. O2 C. NaCl D. CuF2

katrin2010 [14]

O2 is the right answer

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3 years ago

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When the temperature of air rises, the amount of water needed for saturation

rusak2 [61]

It increases as temperature rises.

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3 years ago

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Consider the reaction given below.

Drupady [299]

Answer:

K = 0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.

Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

2ᵃ = 2 ⇒ a = 1

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K = 0.167 s⁻¹

6 0

3 years ago

Which has the highest electronegativity value?

AnnZ [28]

Which has the highest electronegativity value?
A
hydrogen

B
calcium

C
helium

D
fluorine d because fluorine has a higher group number

8 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

3 years ago